 Hi, can someone help Anonymous

### Question Description

A 7.93 kg box is pulled along a horizontal surface by a force FP of 84.0 N applied at a 47.0º angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box? Since,Applied force F = 84 N at an angle of 47 degrees .

The horizontal component of the applied force is
F1 = 84 cos (47) = 84 * 0.682 = 57.288 N
Normal Force = N = m g + F sin 47

= 7.93 * 9.8 + 84 sin (47 )

= 77.71 + 61.43 = 139.143 N
f = µ N = 0.35 * 139.143 = 48.7 N
Net force = F1 - f = 57.288 - 48.7 = 8.588 N
F net = m * a --- a

= Fnet / m

= 8.588 / 7.93

= 1.083 m/s2 Furious (72)
Boston College    Review Anonymous
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